Integraal van $$$4 \sin{\left(\frac{\pi t}{2} \right)}$$$

Bepaal $$$\int 4 \sin{\left(\frac{\pi t}{2} \right)}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=4$$$ en $$$f{\left(t \right)} = \sin{\left(\frac{\pi t}{2} \right)}$$$:

$${\color{red}{\int{4 \sin{\left(\frac{\pi t}{2} \right)} d t}}} = {\color{red}{\left(4 \int{\sin{\left(\frac{\pi t}{2} \right)} d t}\right)}}$$

Zij $$$u=\frac{\pi t}{2}$$$.

Dan $$$du=\left(\frac{\pi t}{2}\right)^{\prime }dt = \frac{\pi}{2} dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$dt = \frac{2 du}{\pi}$$$.

Dus,

$$4 {\color{red}{\int{\sin{\left(\frac{\pi t}{2} \right)} d t}}} = 4 {\color{red}{\int{\frac{2 \sin{\left(u \right)}}{\pi} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{2}{\pi}$$$ en $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$4 {\color{red}{\int{\frac{2 \sin{\left(u \right)}}{\pi} d u}}} = 4 {\color{red}{\left(\frac{2 \int{\sin{\left(u \right)} d u}}{\pi}\right)}}$$

De integraal van de sinus is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$\frac{8 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{\pi} = \frac{8 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{\pi}$$

We herinneren eraan dat $$$u=\frac{\pi t}{2}$$$:

$$- \frac{8 \cos{\left({\color{red}{u}} \right)}}{\pi} = - \frac{8 \cos{\left({\color{red}{\left(\frac{\pi t}{2}\right)}} \right)}}{\pi}$$

Dus,

$$\int{4 \sin{\left(\frac{\pi t}{2} \right)} d t} = - \frac{8 \cos{\left(\frac{\pi t}{2} \right)}}{\pi}$$

Voeg de integratieconstante toe:

$$\int{4 \sin{\left(\frac{\pi t}{2} \right)} d t} = - \frac{8 \cos{\left(\frac{\pi t}{2} \right)}}{\pi}+C$$

$$$\int 4 \sin{\left(\frac{\pi t}{2} \right)}\, dt = - \frac{8 \cos{\left(\frac{\pi t}{2} \right)}}{\pi} + C$$$A